How would you do a proof of 'n' is the sum of two consecutive perfect squares, prove 2n-1 is a square number

Question

'n' is the sum of two consecutive perfect squares, prove 2n-1 is a square number

Answer 1

$$n=a^2+(a+1)^2$$

$$n=2a^2+2a+1$$,

$$2n-1=4a^2+4a+1$$

$$2n-1=(2a+1)^2$$

Answer 2

$$n=k^2+(k+1)^2=2k^2+2k+1$$ $$2n-1=4k^2+4k+1 = (2k+1)^2$$